Is csc(θ)(1-cos^2(θ))=sin(θ) ?

The answer to whether csc(θ)(1-cos^2(θ))=sin(θ) is True

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prove csc(θ)(1-cos^2(θ))=sin(θ)

prove

csc (θ) (1 - cos^{2} (θ)) = sin (θ)

True

Solution steps

csc (θ) (1 - cos^{2} (θ)) = sin (θ)

Manipulating left side

csc (θ) (1 - cos^{2} (θ))

Express with sin, cos

(1 - cos^{2} (θ)) csc (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= (1 - cos^{2} (θ)) \frac{1}{sin ( θ )}

Simplify

(1 - cos^{2} (θ)) \frac{1}{sin ( θ )} : \frac{1 - cos ^{2} ( θ )}{sin ( θ )}

(1 - cos^{2} (θ)) \frac{1}{sin ( θ )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( 1 - cos ^{2} ( θ ) )}{sin ( θ )}

1 \cdot (1 - cos^{2} (θ)) = 1 - cos^{2} (θ)

1 \cdot (1 - cos^{2} (θ))

Multiply:

1 \cdot (1 - cos^{2} (θ)) = (1 - cos^{2} (θ))

= (1 - cos^{2} (θ))

Remove parentheses:

(a) = a

= 1 - cos^{2} (θ)

= \frac{1 - cos ^{2} ( θ )}{sin ( θ )}

= \frac{1 - cos ^{2} ( θ )}{sin ( θ )}

= \frac{1 - cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

\frac{1 - cos ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{( csc ^{2} ( x ) )}{cot ( x )} = csc (x) sec (x)

csc^{2} (x) (tan^{2} (x)) - 1 = tan^{2} (x)

sin (x) cos (x) + sin (3 x) sec (x) = tan (x)

tan (θ) + sec (θ) = \frac{cos ( θ )}{1 - sin ( θ )}

(1 + sec (x)) (1 - cos (x)) = sin (x) tan (x)

Is csc(θ)(1-cos^2(θ))=sin(θ) ?
The answer to whether csc(θ)(1-cos^2(θ))=sin(θ) is True